schange

Author: eahussein

resources/links/.ipynb_checkpoints/Angular resolution-checkpoint.md

@@ -5,17 +5,17 @@
 - https://en.wikipedia.org/wiki/Angular_resolution
 
 ---
-  
+
 #### What is the angular resolution?
 - Any image-forming device like a telescope can be used to recognise small details of an object. Hence, the smaller the angular resolution, the better the formed image as it can focus on smaller and smaller details. 
 
 **_SO THE GOAL BECOMES HOW TO MAKE IMAGES WITH SMALL ANGULAR RESOLUTIONS._**
 
-It seems that in astronomy, there is a relationship between the image resolution $\theta$ and the wavelength $\lambda$ giving by 
-$$ \theta = 1.22 \frac{\lambda} {D} $$
+It seems that in astronomy, there is a relationship between the image resolution $\theta$ and the wavelength
+$\lambda$ giving by $$ \theta = 1.22 \frac{\lambda} {D} $$
 - Where $D$ is the diameter of the telescope.
 - to get better resolution (small $\theta$), we need:
-	- Large lenses $D$
+	- Large lenses ($D$)
 	- shorter wavelength $\lambda$
 - Therefore, as $\lambda$ increases $\theta$ also increases making the image less detailed.
 - Be careful that larger image resolution does not mean better images 

resources/links/Angular resolution.md

@@ -5,17 +5,17 @@
 - https://en.wikipedia.org/wiki/Angular_resolution
 
 ---
-  
+
 #### What is the angular resolution?
 - Any image-forming device like a telescope can be used to recognise small details of an object. Hence, the smaller the angular resolution, the better the formed image as it can focus on smaller and smaller details. 
 
 **_SO THE GOAL BECOMES HOW TO MAKE IMAGES WITH SMALL ANGULAR RESOLUTIONS._**
 
-It seems that in astronomy, there is a relationship between the image resolution $\theta$ and the wavelength $\lambda$ giving by 
-$$ \theta = 1.22 \frac{\lambda} {D} $$
+It seems that in astronomy, there is a relationship between the image resolution $\theta$ and the wavelength
+$\lambda$ giving by $$ \theta = 1.22 \frac{\lambda} {D} $$
 - Where $D$ is the diameter of the telescope.
 - to get better resolution (small $\theta$), we need:
-	- Large lenses $D$
+	- Large lenses ($D$)
 	- shorter wavelength $\lambda$
 - Therefore, as $\lambda$ increases $\theta$ also increases making the image less detailed.
 - Be careful that larger image resolution does not mean better images